Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $p = \dfrac{-q^3 - 9q^2 - 20q}{5q^3 - 20q^2 - 160q} \div \dfrac{2q + 2}{-q + 8} $
Dividing by an expression is the same as multiplying by its inverse. $p = \dfrac{-q^3 - 9q^2 - 20q}{5q^3 - 20q^2 - 160q} \times \dfrac{-q + 8}{2q + 2} $ First factor out any common factors. $p = \dfrac{-q(q^2 + 9q + 20)}{5q(q^2 - 4q - 32)} \times \dfrac{-(q - 8)}{2(q + 1)} $ Then factor the quadratic expressions. $p = \dfrac {-q(q + 4)(q + 5)} {5q(q + 4)(q - 8)} \times \dfrac {-(q - 8)} {2(q + 1)} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac { -q(q + 4)(q + 5) \times -(q - 8)} { 5q(q + 4)(q - 8) \times 2(q + 1)} $ $p = \dfrac {q(q + 4)(q + 5)(q - 8)} {10q(q + 4)(q - 8)(q + 1)} $ Notice that $(q + 4)$ and $(q - 8)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac {q\cancel{(q + 4)}(q + 5)(q - 8)} {10q\cancel{(q + 4)}(q - 8)(q + 1)} $ We are dividing by $q + 4$ , so $q + 4 \neq 0$ Therefore, $q \neq -4$ $p = \dfrac {q\cancel{(q + 4)}(q + 5)\cancel{(q - 8)}} {10q\cancel{(q + 4)}\cancel{(q - 8)}(q + 1)} $ We are dividing by $q - 8$ , so $q - 8 \neq 0$ Therefore, $q \neq 8$ $p = \dfrac {q(q + 5)} {10q(q + 1)} $ $ p = \dfrac{q + 5}{10(q + 1)}; q \neq -4; q \neq 8 $